Question

A cylinder of mass $m$ is kept on the edge of a plank of mass $2m$ and length $12m$, which in turn is kept on smooth ground. Coefficient of friction between the plank and the cylinder is $0.1$. The cylinder is given an impulse, which imparts it a velocity $7$ ${ms}^{-1}$ but no angular velocity. If the time after which the cylinder falls off the plank is $t$ seconds, find $4t$.

Answer 1

friction acts when two surfaces with common interface has different velocities.
here the common points are the bottom most point of cylinder and the corresponding point on the slab.
initially ball is translating, but friction comes into action and starts rotating the ball, at the same time, the friction also accelerates the slab giving it velocity. The friction stops acting when velocity of ball's lowest point and slab are equal.
f = 0.1mg => f/m = 1$m/s^2$
$a_m=1m/s^2;a_{2m}=f/2m=0.5m/s^2$
$\tau =M{R}^2\alpha /2=f\ast r=0.1MgR$
$R\alpha =0.2g=2m/s^2$
Therefore, velocity of P(lowest point on cylinder) = $v_p=7-(2+1)t=7-3t$
velocity of slab $v_l=0.5t$
these equations are apt till friction acts, i.e, till $v_p=v_l$
$7-3t=0.5t=>t=2sec$
$v_l=1m/s,v_{ball}=7-t=5m/s$
total time taken to fall of the slab = time taken to cover distance.
I) when friction acts:
$d_{ball}={\int }_0^{2}7-tdt=14-2=12m$
$d_{slab}={\int }_0^{2}0.5tdt=0.5\ast 2^2/2=1m$
net distance = 11m, remaining 1m. 1m to be travelled when no friction acts.
II) $v_{ball}-v_{slab}=5-1=4m/s$
$t_2=1/4=0.25sec$
Total time taken = 2+0.25 = 2.25sec. 4t = 9.